Let $y=\sqrt{x}e^x$. $\dfrac{dy}{dx}=$
Explanation: $\sqrt{x}e^x$ is the product of two, more basic, expressions: $\sqrt{x}$ and $e^x$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\sqrt{x}e^x\right) \\\\ &=\dfrac{d}{dx}\left(\sqrt{x}\right)e^x+\sqrt{x}\dfrac{d}{dx}(e^x)&&\gray{\text{The product rule}} \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{1}{2}}}\right)e^x+\sqrt{x}\dfrac{d}{dx}(e^x)&&\gray{\text{Write } \sqrt{x}\text{ as a power}} \\\\ &=\dfrac12x^{^{-\frac{1}{2}}}\cdot e^x+\sqrt{x}\cdot e^x&&\gray{\text{Differentiate }x^{^{\frac{1}{2}}}\text{ and }e^x} \\\\ &=e^x\left(\dfrac{1}{2\sqrt{x}}+\sqrt{x}\right)&&\gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=e^x\left(\dfrac{1}{2\sqrt{x}}+\sqrt{x}\right)$ or any other equivalent form.